a) (y – 607 200) : 305 = 642 + 318
c) (1780 – 973) x (75 : y) = 2401 + 20
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a) (y – 1) : 105 = 125 x 80
(y – 1) : 105 = 10000
(y – 1) = 10000 x 105
(y – 1) = 1050000
y = 1050000 + 1
y = 1050001
b) (y – 607 200) : 305 = 642 + 318
(y – 607 200) : 305 = 960
(y – 607 200) = 960 x 305
(y – 607 200) = 292800
y = 292800 + 607 200
y = 900000
\(a.\left(y-1\right):105=125x80\) \(b.\left(y-607200\right):305=642+318\)
\(\left(y-1\right):105=10000\) \(\left(y-607200\right):305=960\)
\(y-1=10000x105=1050000\) \(y=960+607200=608160\)
\(y=1050000+1=1050001\)
\(c.\left(1780-973\right)x\left(75:y\right)=2401+20\)
\(807x\left(75:y\right)=2401+20\)
\(807x\left(75:y\right)=2421\)
\(75:y=2421:807=3\)
\(y=75:3=25\)
a. x : 4 - 11250 = 22850
x : 4 = 22850 + 11250
x = 34100 : 4
x = 8525
b. (x - 60720) : 5 = 318 + 642
x - 60720 = 960 x 5
x = 4800 + 60720
x = 65520
c. (x - 1) : 8 = 1000
x - 1 = 1000 x 8
x = 8000 + 1
x = 8001
Ta có \(\dfrac{a}{3b}=\dfrac{b}{3c}=\dfrac{c}{3d}=\dfrac{d}{3a}\)
Áp dụng tính chất dãy tỉ số bằng nhau
\(\Rightarrow\dfrac{a}{3b}=\dfrac{b}{3c}=\dfrac{c}{3d}=\dfrac{d}{3a}=\dfrac{a+b+c+d}{3\left(a+b+c+d\right)}=\dfrac{1}{3}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{3b}=\dfrac{1}{3}\\\dfrac{b}{3c}=\dfrac{1}{3}\\\dfrac{c}{3d}=\dfrac{1}{3}\\\dfrac{d}{3a}=\dfrac{1}{3}\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}3a=3b\\3b=3c\\3c=3d\\3d=3a\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}a=b\\b=c\\c=d\\d=a\end{matrix}\right.\)
\(\Leftrightarrow a=b=c=d\) ( đpcm )
a) (x + 74) - 318 = 200
x + 74 = 200 + 318
x + 74 = 518
x = 518 - 74
x = 444
b) 3636 : (12x - 91) = 36
12x - 91 = 3636 : 36
12x - 91 = 101
12x = 101 + 91
12x = 192
x = 192 : 12
x = 16
c) (x : 23 + 45) . 67 = 8911
x : 23 + 45 = 8911 : 67
x : 23 + 45 = 133
x : 23 = 133 - 45
x : 23 = 88
x = 88 . 23
x = 2024
a)\(\left(y-607200\right):305=642+318\)
⇔\(\left(y-607200\right):305=960\)
⇔\(y-607200=292800\)
⇔\(y=900000\)
a)(y−607200):305=642+318(y−607200):305=642+318
(y−607200):305=960(y−607200):305=960
y−607200=292800y−607200=292800
y=900000